Shaun Williams, PhD

- Chemical kinetics is the study of chemical reaction rates.
- In order to understand kinetics, there are some background equations that we need.
- These equations will be fully developed later in the course.

- \( \gamma = \rho \sigma \left< v_{AA} \right> \)
- The effective cross-sectional area of the colliding molecules is $$ \sigma_{AB}=\frac{1}{4}\left( \sigma_A + 2\sqrt{\sigma_A \sigma_B} +\sigma_B \right) $$
- The average relative speed is $$ \left< v_{AB} \right> = \sqrt{\frac{8k_\mathrm{B}T}{\pi \mu}} $$
- The reduced mass is \( \mu = \frac{m_A m_B}{m_A+m_B} \)

- Some reaction rates are limited by diffusion
- In these cases, we will use Fick's first law to help quantify the dynamics.
- \( J(Z_0) = -D\left. \left( \frac{d\rho}{dZ} \right) \right|_{Z_0} \)

- The equilibrium constant can be written in terms of partition function (a lot more on that later in the course) $$ K_{\mathrm{eq}} = \frac{Q_{\mathrm{products}}}{Q_{\mathrm{reactants}}} $$
- We can also write it in terms of the standard state Gibbs' Free energy \( \Delta_{\mathrm{rxn}} G^\ominus \) $$ K_\mathrm{eq} = e^{-\frac{\Delta_\mathrm{rxn}G^\ominus}{RT}} $$
- The Gibbs' Free energy can be broken down as \( \Delta G = \Delta H - T \Delta S \)
- The enthalpy for any process is given as \( H = E + PV \)

- Chemical kinetics is the study of chemical reaction rates.
- In kinetics, both microscopic and macroscopic parameters of the system are crucial to understanding chemical behavior

- Consider the reaction $$ \mathrm{Cl(g) + HF(g) \rightarrow HCl(g) + F(g)} $$
- For this reaction $$ \Delta_\mathrm{rxn}H^\ominus = 165.198+173.779-186.908-158.754 = -6.685\,\frac{kJ}{mol} $$
- This reaction is less probable if the chlorine attacks from the F end of the HF molecule.
- This reaction is even less probable if the fluorine atom is replaced wiht much larger functional groups. $$ \mathrm{Cl(g) + (CF_3)_3CH(g) \rightarrow HCl(g) + (CF_3)_3C(g)} $$
- Because of these steric considerations we should expect that the first reaction will be much more probable than the second reaction
- This means that we should expect the reaction rate of the first reaction to be faster than the reaction rate of the second reaction.

- During a chemical reaction, the potential energy of the system tends to increase.
- The total energy of the system, however, is constant.
- Basically, the reacting system is exchanging kinetic energy (vibrational and rotational) for potential energy.
- A chemical reaction must overcome the activation energy in order to react.
- The activation energy is the peak in potential energy between the reactants and products.
- The height of this barrier depends on the molecular structure and quantum mechanics.

- The rate at which we observe changes in a system depends on the system's thermodynamic parameters.

- The language of chemical kinetics grew out of the study of reactions in solutions.
- We tend to represent the amount of each molecule in formulas using molarity.
- Even in gas phase reactions we describe the concentration using the number density is particles per \( \mathrm{cm}^{-3} \)

- The rate constant is a function of temperature (we will later determine this dependence)

- We can define a reaction velocity as $$ v=\frac{1}{\nu_i}\frac{d\left[ i \right]}{dt} = \frac{1}{V\nu_i}\frac{dn_i}{dt} $$
- \( \left[ i \right] \) - molar concentation of product or reactant \( i \)
- \( \nu_i \) - the stoichiometric coefficient of \( i \) in the reaction
- \( n_i \) - the number of moles of component \( i \) contained in volume \( V \)
- The use of the stoiciometric coefficient in the equation is to ensure that the same reaction rate is found regardless of what component is being considered.
- For the reaction \( \mathrm{2H_2 + O_2 \rightarrow 2H_2O} \) $$ -\frac{1}{2}\frac{d\left[ \mathrm{H_2}\right]}{dt} = -\frac{d\left[ \mathrm{O_2}\right]}{dt} = \frac{1}{2}\frac{d\left[ \mathrm{H_2O}\right]}{dt} $$

- Often, the reaction velocity is written in terms of a rate law
- The rate law is a power law in the reactant concentrations
- The rate law includes a concentration-independent coefficient called the rate constant, \(k \)
- An example of a generic rate law $$ v=k\left[ A\right]^x\left[ B\right]^y\dots $$
- The reaction order is the sum of the exponents, \( x+y+\dots \)

- A net rreaction is a reaction written to show only the initial and final chemical components, for example $$ \mathrm{2H_2 + O_2 \rightarrow 2H_2O} $$
- This reaction actually occurs in a series of steps $$ \begin{align} \mathrm{H_2 + O_2} & \rightarrow \mathrm{H+HO_2} \\ \mathrm{HO_2 + H_2} & \rightarrow \mathrm{H_2O + OH} \\ \mathrm{H + OH + M} & \rightarrow \mathrm{H_2O + M} \end{align} $$ The \( M \) in the last step is shorthand for any particle that participates in the reaction without being changed
- Each of these individual steps is called an elementary step.
- Elementary steps require only a single collision or decay.

- The reaction mechanism is the series of elementary reaction that lead to a net reaction equation
- On the previous slide, if we add up the three elementary steps, we get the net equation
- In this chapter we will focus on elementary reactions

- If the initial reactant is unstable $$ \mathrm{XeCl \rightarrow Xe + Cl} $$
- If the reactant molecule has been previous energized (through collisions or laser) $$ \mathrm{C_2H_4O_2\,(dioxetane) \rightarrow 2H_2CO} $$

- A bimolecular elementary reaction involves the collision between two reactants $$ \mathrm{Cl + HCl \rightarrow Cl_2 + H} $$
- A termolecular elementary reaction involves the collision between three reactants $$ \mathrm{He + H + Cl \rightarrow He + HCl} $$
- It is important to note that the probability of three reactants colliding simultaneously is very small compared to bimolecular collisions
- Elementary reactions involving more than three reactants have not been found.

- We now what to fold together the parameters to predict the value of the rate constant, \( k \)
- For our simplest model of chemical reactions, we fix the temperature and assume the following
- The reaction is bimolecular, \( \mathrm{A + B} \xrightarrow{k} \mathrm{products} \)
- The reactants must be close enough for intermolecular forces to be important before the reaction can take place.
- A single distribution function describes the probability of the reaction occuring
- There will be some activation barrier to overcome

- First we count the number of collisions per unit time
- Multiply that by the fraction of those collision with enough kinetic energy
- this bring in the canonical factor \( e^{-\frac{E_a}{RT}} \) (later in the course)
- Bringing all this together we arrive at a version of the empirical Arrhenius equation $$ k_\mathrm{Arr}=Ae^{-\frac{E_a}{RT}} $$ namely $$ k_\mathrm{SCT} = \sigma_\mathrm{AB} \sqrt{\frac{8k_\mathrm{B}T}{\pi \mu}} \mathcal{N}_\mathrm{A} p e^{-\frac{E_a}{RT}} $$ where \( p \) is a steric factor.

- Note: When \( RT=E_a \), \( k_\mathrm{Arr} \) is about at one-third of its maximum value.

Estimate the value of \( A \) in \( \mathrm{cm^3\,mol^{-1}\,s^{-1}} \) for the reaction between oxygen atom and molecular nitrogen, \( \mathrm{O(g) + N_2(g)} \), at \( 298\,\mathrm{K} \). Let the collision cross-section be \( 30\, \AA^2 \) for oxygen and \( 37\, \AA^2 \) for \( \mathrm{N_2} \).

- We would like to be able to measure the value of \( k \) at various temperatures in the lab and determine the value of \( E_a \)
- To do this, we first need to rearrange the Arrhenius equation

$$ \begin{align} k & = A \exp \left( -\frac{E_a}{RT} \right) \\ \frac{k}{A} & = \exp \left( -\frac{E_a}{RT} \right) \\ \ln \frac{k}{A} & = -\frac{E_a}{RT} \end{align} $$

$$ \begin{align} \frac{d}{dT} \left( \ln \frac{k}{A} \right) & = -\frac{d}{dT} \frac{E_a}{RT} \\ \frac{d}{dT} \left( \ln k - \ln A \right) & = \frac{d}{dT} \ln k = -\left( -\frac{E_a}{RT^2} \right) = \frac{E_a}{RT^2} \\ E_a & \equiv RT^2 \frac{d\ln k}{dT} \end{align} $$ This result is great, but experimentally, we needs a more useful approach.

Let's try a more useful approach $$ \begin{align} \ln \frac{k}{A} & = -\frac{E_a}{RT} \\ \ln k - \ln A & = -\frac{E_a}{RT} \\ \ln k & = -\frac{E_a}{R}\frac{1}{T} + \ln A \end{align} $$ This is now linear. A plot of \( \ln k \) versus \( \frac{1}{T} \) will provide a straight line with a slope of \( -\frac{E_a}{R} \) and a y-intercept of \( \ln A \).

For the reaction \( 2NOBr \rightarrow 2NO + Br_2 \), calculate \( k \) at \( 298\,\mathrm{K} \) assuming only the Arrhenius equation (constants given in Table 13.1 of the book). Then predict the values of \( k \) at \( 400\,\mathrm{K} \) if (a) the Arrhenius equation continues to hold and (b) if the Arrhenius value at \( 298\,\mathrm{K} \) is correct but \( A \) is actually of the form \( A' T^\frac{1}{2} \).

- Many kinetic models take advantage of transition state theory or activated complex theory
- This theory describes a chemical reaction as a process in which reactants combine to form an activated complex which then decays to product $$ \mathrm{A+B \rightarrow AB^\ddagger \rightarrow products} $$

- The rate constant equation that we get utilizing transition state theory is called the Eyring equation $$ k_\mathrm{TST} = \frac{k_\mathrm{B}T}{Ch}\exp \left( \frac{\Delta S^\ddagger}{R} \right) \exp \left( -\frac{\Delta H^\ddagger}{RT} \right) $$ where \( C \) is the overall concentration
- Here we have a slightly more complex version than Arrhenius involving more thermodynamic quantities
- Our steric factor has been replaced by a term dependent on the entropy change which makes sense

- Our equations thus far have only taken into account gas phase diffusion
- In order to correctly model reactions in solutions we need to allow for the diffusion of the reactants in a solvent
- We are going to use Fick's first law to do this $$ J=-D\left( \frac{d\rho}{dZ} \right) $$ where \( J \) is the diffusion-limited flux of molecules per unit area per unit time

- To estimate the diffusion-limited rate constant, we assume a sphere centered on each molecule A such that any molecule B that gets inside the sphere will react.

- The rate, in molecules per unit time, of B diffusing into any sphere of radius \( r \) is given the symbol \( \gamma_\mathrm{D} \) $$ \begin{align} \gamma_\mathrm{D} & = -J_\mathrm{B} \left( 4\pi r^2 \right) \\ & = D_\mathrm{AB} \left( 4\pi r^2 \right) \left( \frac{d\rho_\mathrm{B}}{dr} \right) \end{align} $$
- Doing some math we arrive at $$ \gamma_\mathrm{D} = 4\pi R_\mathrm{AB} D_\mathrm{AB} \rho_\mathrm{B} $$

- Using our diffusion rate equation we can derive a rate constant
- This rate constant assumes:
- Diffusion-limited rate
- A bimolecular reaction
- The diffusion-limited rate constant is $$ k_\mathrm{D}=4\pi R_\mathrm{AB} D_\mathrm{AB} $$

- Both simple collision and transition state theories find that the reaction rate depends solely on the rate constant and the concentrations of the reactants $$ \mathrm{elementary\, reaction:}\, v=k\prod_\mathrm{reactants} \left[ \mathrm{i} \right]^{v_\mathrm{i}} $$
- Some examples: $$ \begin{align} \mathrm{A \rightarrow products:}\; & v=k\left[ \mathrm{A} \right] \\ \mathrm{A + B \rightarrow products:}\; & v=k\left[ \mathrm{A} \right] \left[ \mathrm{B} \right] \\ \mathrm{2A \rightarrow products:}\; & v=k\left[ \mathrm{A} \right]^2 \\ \mathrm{A + B + C \rightarrow products:}\; & v=k\left[ \mathrm{A} \right] \left[ \mathrm{B} \right] \left[ \mathrm{C} \right] \\ \end{align} $$

- Let's look at a generic unimolecular decomposition reaction, \( A \rightarrow B + C \)
- The rate law for this reaction is $$ v = \frac{d\left[\mathrm{B}\right]}{dt} = -\frac{d\left[\mathrm{A}\right]}{dt} = k\left[\mathrm{A}\right] $$
- We would like to come up with an equation that allows us to determine the concentration of a species as a function of time
- To do this, we need to do a little math

- Starting with \( -\frac{d\left[\mathrm{A}\right]}{dt} = k\left[\mathrm{A}\right] \) we find that: $$ \begin{align} -\frac{d\left[\mathrm{A}\right]}{\left[\mathrm{A}\right]} & = k\, dt \\ -\int_{\left[\mathrm{A}\right]_0}^{\left[\mathrm{A}\right]} \frac{d\left[\mathrm{A}\right]'}{\left[\mathrm{A}\right]'} & = k\int_0^t dt \\ -\ln \frac{\left[\mathrm{A}\right]}{\left[\mathrm{A}\right]_0} & = kt \; \therefore \; \left[\mathrm{A}\right] = \left[\mathrm{A}\right]_0 e^{-kt} \\ \mathrm{if\, \left[B\right]_0 = \left[C\right]_0 = 0:}\; \left[\mathrm{B}\right] & = \left[\mathrm{C}\right] = \left[\mathrm{A}\right]_0\left(1-e^{-kt}\right) \end{align} $$

- Once we have the integrated rate law we can determine the reaction half-life
- The half-life, \( t_\bfrac{1}{2} \), is the time is takes the reaction for \( \left[\mathrm{A}\right] = \frac{1}{2}\left[\mathrm{A}\right]_0 \)
- So $$ \begin{align} -\ln \frac{\left[\mathrm{A}\right]}{\left[\mathrm{A}\right]_0} & = kt \\ -\ln \frac{\frac{1}{2}\left[\mathrm{A}\right]_0}{\left[\mathrm{A}\right]_0} & = kt_\bfrac{1}{2} \\ -\ln \frac{1}{2} & = kt_\bfrac{1}{2} \; \therefore \; t_\bfrac{1}{2} = -\frac{1}{k}\ln \frac{1}{2} = \frac{\ln 2}{k} \end{align} $$

Find the integrated rate law for the elementary reaction \( \mathrm{2A\rightarrow B} \), where \( \mathrm{A} \) is butadiene and \( \mathrm{B} \) is vinylcyclohexene. Use this to obtain an expression for the half-life of butadiene in terms of the rate constant \( k \) and the initial concentration \( \left[ \mathrm{A} \right]_0 \).

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